POJ 2891（线性同余方程组的求解）

最新推荐文章于 2022-02-03 23:23:34 发布

spring371327

最新推荐文章于 2022-02-03 23:23:34 发布

阅读量806

点赞数

CC 4.0 BY-SA版权

分类专栏：数论文章标签： POJ 2891 中国剩余定理线性同余方程组的求解

本文链接：https://blog.youkuaiyun.com/spring371327/article/details/52735528

数论专栏收录该内容

17 篇文章

订阅专栏

Strange Way to Express Integers

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 14494		Accepted: 4743

Description

Elina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative integers. The way is described as following:

Choose k different positive integers a₁, a₂, …, a_k. For some non-negative m, divide it by every a_i (1 ≤ i ≤ k) to find the remainder r_i. If a₁, a₂, …, a_k are properly chosen, m can be determined, then the pairs (a_i, r_i) can be used to express m.

“It is easy to calculate the pairs from m, ” said Elina. “But how can I find m from the pairs?”

Since Elina is new to programming, this problem is too difficult for her. Can you help her?

Input

The input contains multiple test cases. Each test cases consists of some lines.

Line 1: Contains the integer k.
Lines 2 ~ k + 1: Each contains a pair of integers a_i, r_i (1 ≤ i ≤ k).

Output

Output the non-negative integer m on a separate line for each test case. If there are multiple possible values, output the smallest one. If there are no possible values, output -1.

Sample Input

2
8 7
11 9

Sample Output

Hint

All integers in the input and the output are non-negative and can be represented by 64-bit integral types.

思路：

线性同余方程组的求解，采用递归式的两两求解方法。无数论基础的自行百度该类方程的解法，内容太多，懒得写了。。。

代码：

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <iostream>
using namespace std;

typedef long long int ll;


ll ext_gcd(ll a,ll b,ll& s, ll& t){
	if(b==0){
		s = 1;
		t = 0;
		return a;
	}
	ll ans = ext_gcd(b,a%b,s,t);
	ll tmp = s;
	s = t;
	t = tmp - a/b*t;
	return ans;
}

int main(){
	
	ll N;
	ll m,r;
	ll R,M;
	ll s,t;
	ll gcd;
	ll i;
	ll tmp;
	int flag = 0; 
	while(~scanf("%I64d",&N)){
		flag = 0;
		scanf("%I64d%I64d",&M,&R);
		for(i=2; i<=N; i++){
			scanf("%I64d%I64d",&m,&r);
			gcd = ext_gcd(M,m,s,t);
			if( ((r-R)%gcd)!=0 )
				flag = 1;		
			//这个地方一定要注意不要break出来，因为break出来，有些数据没读，被当成下一组数据的输入
			//会出现TLE和RW等错误的，调了半个小时，血的教训 
			tmp = ((r-R)/gcd*s)%(m/gcd); //不定方程的最小解（不一定正的） 
			R = tmp*M+R;
			M = M/gcd*m;
			R = ( (R%M)+M )%M;
		}
		if(flag)
			printf("-1\n");
		else
			printf("%I64d\n",R);
	}
	return 0;
}